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iogann1982 [59]
4 years ago
5

A gas at 325 k has a volume of 4.0 L. If the gas’s volume changes to 2.0 L at constant pressure, what will be the new temperatur

e
Chemistry
2 answers:
inessss [21]4 years ago
8 0
4/325 = 2/unknown temperature
unknown temperature= 2/(4/325)=162.5k
eimsori [14]4 years ago
6 0
I'm assuming this can be solved by cross multiplication. Sorry if I'm wrong.
325/4.0 x/2.0
x=160
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andriy [413]

Answer:

ionic

Explanation:

4 0
3 years ago
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How many hours will it take for the concentration of methyl isonitrile to drop to 15.0 % of its initial value?
SpyIntel [72]

Answer:

The concentration  of methyl isonitrile will become 15% of the initial value after 10.31 hrs.

Explanation:

As the data the rate constant is not given in this description, However from observing the complete question  the rate constant is given as a rate constant of 5.11x10-5s-1 at 472k .

Now the ratio of two concentrations is given as

ln (\frac{C}{C_0})=-kt

Here C/C_0 is the ratio of concentration which is given as 15% or 0.15.

k is the rate constant which is given as 5.11 \times 10^{-5} \, s^{-1}

So time t is given as

ln (\frac{C}{C_0})=-kt\\ln(0.15)=-5.11 \times 10^{-5} \times t\\t=\frac{ln(0.15)}{-5.11 \times 10^{-5} }\\t=37125.6 s\\t=37125.6/3600 \\t= 10.31 \, hrs

So the concentration will become 15% of the initial value after 10.31 hrs.

7 0
3 years ago
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A saturated solution is made by dissolving 0.327 g of a polypeptide (a substance formed by joining together in a chainlike fashi
faust18 [17]

Answer: The approximate molecular mass of the polypeptide is 856 g/mol

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

Or,

\pi=i\times \frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution in L)}}\times RT

where,

\pi = osmotic pressure of the solution = 4.19 torr

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (polypeptide) = 0.327 g

Volume of solution = 1.70 L

R = Gas constant = 62.364\text{ L.torr }mol^{-1}K^{-1}

T = temperature of the solution = 26^oC=[273+26]K=299K

Putting values in above equation, we get:

4.19torr=1\times \frac{0.327}{\text{Molar mass of solute}\times 1.70}\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 299K\\\\\text{molar mass of solute}=856g/mol

Hence, the molar mass of the polypeptide is 856 g/mol

8 0
3 years ago
Liquid Q is a polar solvent and liquid R is a nonpolar solvent. On the basis of this information, you would expect:
bearhunter [10]
4) is correct
This is because water is polar and it will mix with a polar solvent. A good rule for remembering the behavior of non-polar and polar compounds when it comes to being miscible is that "like dissolves like."
6 0
4 years ago
A 0.450 g sample of solid lead(II) nitrate is added to 250 mL of 0.250 M sodium iodide solution. Assume no change in volume of t
Verdich [7]

Pb(NO₃)₂ ⇒limiting reactant

moles PbI₂ = 1.36 x 10⁻³

% yield  = 87.72%

<h3>Further explanation</h3>

Given

Reaction(unbalanced)

Pb(NO₃)₂(s) + NaI(aq) → PbI₂(s) + NaNO₃(aq)

Required

  • moles of PbI₂
  • Limiting reactant
  • % yield

Solution

Balanced equation :

Pb(NO₃)₂(s) + 2NaI(aq) → PbI₂(s) + 2NaNO₃(aq)

mol Pb(NO₃)₂ :

= 0.45 : 331 g/mol

= 1.36 x 10⁻³

mol NaI :

= 250 ml x 0.25 M

= 0.0625

Limiting reactant (mol : coefficient)

Pb(NO₃)₂ : 1.36 x 10⁻³ : 1 = 1.36 x 10⁻³

NaI : 0.0625 : 2 = 0.03125

Pb(NO₃)₂ ⇒limiting reactant(smaller ratio)

moles PbI₂ = moles Pb(NO₃)₂ = 1.36 x 10⁻³(mol ratio 1 : 1)

Mass of PbI₂ :

= mol x MW

=  1.36 x 10⁻³ x 461,01 g/mol

= 0.627 g

% yield = 0.55/0.627 x 100% = 87.72%

7 0
3 years ago
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